The WKB approximation states that in one dimension, the tunneling probability $P$ can be approximated as
$\ln P=-\frac{2\sqrt{2m}}{\hbar}\int_a^b \sqrt{V-E} dx$ ,
where the limits of integration $a$ and $b$ are the classical turning points, $m$ is the reduced mass, the electrical potential $V$ is a function of $x$, and $E$ is the total energy. Setting $V=kq_1q_2/x$, we have for the integral
$I=\int_a^b \sqrt{V-E} dx$
$=\frac{kq_1k_2}{\sqrt{E}}\int_{A}^1\sqrt{u^{-1}-1} du$ ,
where $A=a/b$. The indefinite integral equals $-u\sqrt{u^{-1}-1}+\tan^{-1}\sqrt{u^{-1}-1}$, and for $A\ll 1$ the definite integral is then $\pi/2$. The result is
$\ln P=-\frac{\pi kq_1q_2}{\hbar}\sqrt{\frac{2m}{E}} $ .
This result was obtained in Gamow 1938, and $G=-(1/2)\ln P$ is referred to as the Gamow factor or Gamow-Sommerfeld factor.
The fact that the integral $\int_A^1\ldots$ can be approximated as $\int_0^1\ldots$ tells us that the right-hand tail of the barrier dominates, i.e., it is hard for the nuclei to travel through the very long stretch of $\sim 1$ nm over which the motion is only mildly classically forbidden, but if they can do that, it's relatively easy for them to penetrate the highly classically forbidden region at $x\sim1$ fm. Surprisingly, the result can be written in a form that depends only on $m$ and $E$, but not on $a$, i.e., we don't even have to know the range of the strong nuclear force in order to calculate the result.
The generic WKB expression depends on $E$ through an expression of the form $V-E$, which might have led us to believe that with a 1 MeV barrier, it would make little difference whether $E$ was 1 eV or 1 keV, and fusion would be just as likely in trees and houses as in the sun. But because the tunneling probability is dominated by the tail of the barrier, not its peak, the final result ends up depending on $1/\sqrt{E}$.
Because $P$ increases extremely rapidly as a function of $E$, fusion is dominated by nuclei whose energies lie in the tails of the Maxwellian distribution. There is a narrow range of energies, known as the Gamow window, in which the product of $P$ and the Maxwell distribution is large enough to contibute significantly to the rate of fusion.
Gamow and Teller, Phys. Rev. 53 (1938) 608
Oh, but we do! I'm assuming you mean using the fields to simply collide particles with each other, right? Then that's already being done. For example, take this neat little machine: http://en.wikipedia.org/wiki/Fusor
This one runs on the exact same principle you described (though I'm not quite familiar with the inner workings of the LHC).
For energy generation, most current approaches to nuclear fusion here in Europe heat up those exact particles to high temperatures, making them collide into each other due to their high kinetic energies, and then use magnetic fields to steer the particles (en masse in the form of plasma) to keep it confined, to keep it from running into the walls, damaging them and cooling themselves down. It's really hard to execute in practice for now, but we're getting there with projects like JET and ITER. Here you go for some further reading: http://en.wikipedia.org/wiki/Magnetic_confinement_fusion
Best Answer
Think of 2 hydrogen atoms or, protons more accurately since at those temperature the atoms don't have electrons, it's more of a soup. So, 2 protons, both positively charged so they repel each other, crash into each other pretty rarely, cause it's still a lot of empty space, but they do make contact every so often. The energy required to get 2 protons or any other atomic nuclei to touch is very high. This is called the Coulomb barrier
That's basically all fusion is, it's when 2 atomic nuclei, which naturally repel each other, get pushed close enough to touch and fuse into 1 nuclei.
Once they touch, in the case of protons, then it's a matter of what quantum combination is most likely to follow. Protons actually don't like each other so much more often than not, they'll just say "lets not do this" to each other and they effectively bounce off each other, basically staying hydrogen.
About one time in a million . . . or so, the protons will stay and fuse but for this to happen, one of them has to become a neutron, because 2 protons aren't a stable nucleus. So in this rare occurrence when they do fuse, one proton kicks out a positron, a neutrino and a gamma ray, essentially getting the energy to do this from the fusion and you're left with a proton-neutron bound together (Deuterium) from a proton-proton collision. This happens rarely but because the sun is so large and there's trillions and trillions of proton-proton collisions every second, you get trillions and trillions of Deuterium nuclei formed every second and Deuterium, unlike protons, is very eager to merge with a proton or another Deuterium so, from there, the process continues.
This is also why hydrogen bombs are made with Deuterium, not hydrogen - hydrogen is much much harder to create fusion with.
Because of this, the sun effectively burns very very slowly, but all that's needed is sufficient temperature and pressure to break the coulomb barrier.